Question: Let $g(x)=2x^4-x^3+6x^2$. Find $g'(1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $7$ (Choice C) C $13$ (Choice D) D $17$
Answer: Let's first find the expression for $g'(x)$ and then evaluate it at $x=1$. According to the sum rule, the derivative of $2x^4-x^3+6x^2$ is the sum of the derivatives of $2x^4$, $-x^3$, and $6x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(2x^4)&=2\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=2\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=8x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(2x^4-x^3+6x^2) \\\\ &=2\dfrac{d}{dx}(x^4)-\dfrac{d}{dx}(x^3)+6\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=2\cdot 4x^3-3x^2+6\cdot 2x&&\gray{\text{The power rule}} \\\\ &=8x^3-3x^2+12x \end{aligned}$ So we found that $g'(x)=8x^3-3x^2+12x$. Plugging in $x=1$ and evaluating using the calculator, we find that $g'(1)=17$. In conclusion, $g'(1)=17$.